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The work function of caesium metal is $2.14 \;eV$. When light of frequency $ 6 \times 10^{14}Hz$ is incident on the metal surface , photoemmision of electron occurs. What is the stopping potential of the emitted photoelectrons ?

$\begin {array} {1 1} (a)\;0.345eV & \quad (b)\;0.345V \\ (c)\;03.45V & \quad (d)\;0.03454V \end {array}$

 

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For stopping potential $V_{\circ}$, we can write the equation for kinetic energy as :
$ K = eV_{\circ}$
So, $ V_{\circ}=K/e$
$ = (0.345 \times 1.6 \times 10^{-19} ) / ( 1.6 \times 10^{-19} ) = 0.345V$
Ans : (b)

 

answered Dec 23, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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